In calculus, the chain rule is a formula for the derivative of the composite of two functions.

Explanation

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In intuitive terms, if a variable, y, depends on a second variable, u, which in turn depends on a third variable, x, then the rate of change of y with respect to x can be computed as the product of the rate of change of y with respect to u multiplied by the rate of change of u with respect to x.

The chain rule may be stated in any of several equivalent forms:

$(f\; \backslash circ\; g)\text{'}(x)\; =\; (f(g(x)))\text{'}\; =\; f\text{'}(g(x))\; g\text{'}(x),\backslash ,$

or in the Leibniz notation

$\backslash frac\; \{df\}\{dx\}\; =\; \backslash frac\; \{df\}\; \{dg\}\; \backslash cdot\; \backslash frac\; \{dg\}\{dx\},$

In integration, the counterpart to the chain rule is the substitution rule.

Example

Suppose, for example, that one is climbing a mountain at a rate of 0.5 kilometres per hour. The temperature is lower at higher elevations; suppose the rate by which it decreases is 6 °C per kilometre. If one multiplies 6 °C per kilometre by 0.5 kilometre per hour, one obtains 3 °C per hour. This calculation is a typical chain rule application.

The general power rule

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The general power rule (GPR) is derivable via the chain rule.

Example I

Consider $f(x)\; =\; (x^2\; +\; 1)^3$. $f(x)$ is comparable to $h(g(x))$ where $g(x)\; =\; x^2\; +\; 1$ and $h(x)\; =\; x^3$; thus,

{| $f\; \text{'}(x)\; \backslash ,$ $=\; 3(x^2\; +\; 1)^2(2x)\; \backslash ,$ $=\; 6x(x^2\; +\; 1)^2\; \backslash ,$

Example II

In order to differentiate the trigonometric function

$f(x)\; =\; \backslash sin(x^2),\backslash ,$

one can write $f(x)\; =\; h(g(x))$ with $h(x)\; =\; \backslash sin\; x$ and $g(x)\; =\; x^2$. The chain rule then yields

$f\text{'}(x)\; =\; 2x\; \backslash cos(x^2)\; \backslash ,$

since $h\text{'}(g(x))\; =\; \backslash cos\; (x^2)$ and $g\text{'}(x)\; =\; 2x$.

Chain rule for several variables

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The chain rule works for functions of several variables as well. For example, if we have a function $f(u(x,\; y),\; v(x,\; y))$ where

$u(x,\; y)\; =\; 3x\; +\; y^2\; \backslash mbox\{\; and\; \}\; v(x,\; y)\; =\; \backslash sin(xy)\; \backslash mbox\{\; and\; \}\; f=u+v,\backslash ,$

then

$\{\backslash partial\; f\; \backslash over\; \backslash partial\; x\}=\{\backslash partial\; f\; \backslash over\; \backslash partial\; u\}\{\backslash partial\; u\; \backslash over\; \backslash partial\; x\}+\{\backslash partial\; f\; \backslash over\; \backslash partial\; v\}\{\backslash partial\; v\; \backslash over\; \backslash partial\; x\}=3\; +\; \backslash cos(xy)y.$

Proof of the chain rule

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Let f and g be functions and let x be a number such that f is differentiable at g(x) and g is differentiable at x. Then by the definition of differentiability,

$g(x+\backslash delta)-g(x)=\; \backslash delta\; g\text{'}(x)\; +\; \backslash epsilon(\backslash delta)\; \backslash ,$ where $\backslash frac\{\backslash epsilon(\backslash delta)\}\{\backslash delta\}\; \backslash to\; 0\; \backslash ,$ as $\backslash delta\backslash to\; 0.$

Similarly,

$f(g(x)+\backslash alpha)\; -\; f(g(x))\; =\; \backslash alpha\; f\text{'}(g(x))\; +\; \backslash eta(\backslash alpha)\; \backslash ,$ where $\backslash frac\{\backslash eta(\backslash alpha)\}\{\backslash alpha\}\; \backslash to\; 0\; \backslash ,$ as $\backslash alpha\backslash to\; 0.\; \backslash ,$

Now

{| $f(g(x+\backslash delta))-f(g(x))\backslash ,$ $=\; f(g(x)\; +\; \backslash delta\; g\text{'}(x)+\backslash epsilon(\backslash delta))\; -\; f(g(x))\; \backslash ,$ $=\; \backslash alpha\_\backslash delta\; f\text{'}(g(x))\; +\; \backslash eta(\backslash alpha\_\backslash delta)\; \backslash ,$

where $\backslash alpha\_\backslash delta\; =\; \backslash delta\; g\text{'}(x)\; +\; \backslash epsilon(\backslash delta)\; \backslash ,$. Observe that as $\backslash delta\backslash to\; 0,$ $\backslash frac\{\backslash alpha\_\backslash delta\}\{\backslash delta\}\backslash to\; g\text{'}(x)$ and $\backslash frac\{\backslash eta(\backslash alpha\_\backslash delta)\}\{\backslash delta\}\backslash to\; 0$. Hence

$\backslash frac\{f(g(x+\backslash delta))-f(g(x))\}\{\backslash delta\}\; \backslash to\; g\text{'}(x)f\text{'}(g(x))\backslash mbox\{\; as\; \}\; \backslash delta\; \backslash to\; 0.$

The fundamental chain rule

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The chain rule is a fundamental property of all definitions of derivative and is therefore valid in much more general contexts. For instance, if E, F and G are Banach spaces (which includes Euclidean space) and f : E → F and g : F → G are functions, and if x is an element of E such that f is differentiable at x and g is differentiable at f(x), then the derivative (the Fréchet derivative) of the composition g o f at the point x is given by

$\backslash mbox\{D\}\_x\backslash left(g\; \backslash circ\; f\backslash right)\; =\; \backslash mbox\{D\}\_\{f\backslash left(x\backslash right)\}\backslash left(g\backslash right)\; \backslash circ\; \backslash mbox\{D\}\_x\backslash left(f\backslash right).$

Note that the derivatives here are linear maps and not numbers. If the linear maps are represented as matrices (namely Jacobians), the composition on the right hand side turns into a matrix multiplication.

A particularly clear formulation of the chain rule can be achieved in the most general setting: let M, N and P be C^k manifolds (or even Banach-manifolds) and let

f : M → N and g : N → P

be differentiable maps. The derivative of f, denoted by df, is then a map from the tangent bundle of M to the tangent bundle of N, and we may write

$\backslash mbox\{d\}\backslash left(g\; \backslash circ\; f\backslash right)\; =\; \backslash mbox\{d\}g\; \backslash circ\; \backslash mbox\{d\}f.$

In this way, the formation of derivatives and tangent bundles is seen as a functor on the category of C^∞ manifolds with C^∞ maps as morphisms.

Tensors and the chain rule

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See tensor field for an advanced explanation of the fundamental role the chain rule plays in the geometric nature of tensors.

Higher derivatives

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Faà di Bruno's formula generalizes the chain rule to higher derivatives. The first few derivatives are

$\backslash frac\{df\}\{dx\}\; =\; \backslash frac\{dg\}\{dx\}\; \backslash frac\{df\}\{dg\}$

$$

\frac{d^2 f}{d x^2} = \left(\frac{dg}{dx}\right)^2 \frac{d^2 f}{d g^2} + \frac{d^2 g}{dx^2}\frac{df}{dg}

$$

\frac{d^3 f}{d x^3} = \left(\frac{dg}{dx}\right)^3 \frac{d^3 f}{d g^3} + 3\frac{dg}{dx}\frac{d^2 g}{d x^2} \frac{d^2 f}{d g^2} + \frac{d^3 g}{d x^3} \frac{df}{dg}

$$

\frac{d^4 f}{d x^4} = \left(\frac{dg}{dx}\right)^4 \frac{d^4 f}{dg^4} + 6 \left(\frac{dg}{dx}\right)^2 \frac{d^2 g}{d x^2} \frac{d^3 f}{d g^3} + \left\{ 4 \frac{dg}{dx} \frac{d^3 g}{dx^3} + 3\left(\frac{d^2 g}{dx^2}\right)^2\right\} \frac{d^2 f}{d g^2} + \frac{d^4 g}{dx^4} \frac{df}{dg}

See also

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• Inverse chain rule

Differential calculus

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